Question

Potential along x-axis varies parabolically as shown in the figure. Then electric field at x = 0.25 m will be ________ V/m.

Answer 1

15

Answer 2

The potential along the x-axis varies parabolically as shown in the figure. From the figure, the parabola passes through the origin $(0, 0)$ and has a minimum at $x = 1 \, m$ with a value of $V = -10 \, V$.

The general equation of a parabola with vertex at $(h, k)$ is $V(x) = a(x-h)^2 + k$.

In this case, the vertex is at $(1, -10)$, so $h = 1$ and $k = -10$.

The equation of the parabola is $V(x) = a(x-1)^2 - 10$.

The parabola passes through the origin $(0, 0)$, so substitute $x = 0$ and $V = 0$: $0 = a(0-1)^2 - 10$ $0 = a(1)^2 - 10$ $0 = a - 10$ $a = 10$

So, the equation of the potential is $V(x) = 10(x-1)^2 - 10$.

Expanding this, we get $V(x) = 10(x^2 - 2x + 1) - 10 = 10x^2 - 20x + 10 - 10 = 10x^2 - 20x$.

The electric field along the x-axis is given by $E_x = -\frac{dV}{dx}$.

Differentiating $V(x)$ with respect to $x$: $\frac{dV}{dx} = \frac{d}{dx}(10x^2 - 20x) = 20x - 20$.

Now, we find the electric field at $x = 0.25 \, m$: $E_x(0.25) = -\frac{dV}{dx}\Big|_{x=0.25} = -(20(0.25) - 20)$ $E_x(0.25) = -(5 - 20)$ $E_x(0.25) = -(-15)$ $E_x(0.25) = 15 \, V/m$.

The electric field at $x = 0.25 \, m$ is $15 \, V/m$.